Question

Hi i need some help on question 11 b and c. I have already done a

Answer 1

a. 16.27 cm^3

b. 4.3 cm

c. 325.4 seconds

Explanation:

The first thing is to calculate the value of the volume, which is the sum of the volume of each part, like this:

`\begin{gathered} V=V_t+V_c+V_s \\ r=(d)/(2)=(2.6)/(2)=1.3 \\ V_t=A_b\cdot(h)/(3)=\pi\cdot(r)^2\cdot(h)/(3)=3.14\cdot(1.3)^2\cdot(1.2)/(3)=2.12cm^3 \\ V_c=A_b\cdot h=\pi\cdot(r)^2\cdot h=3.14\cdot(1.3)^2\cdot1.8=9.55m^3 \\ V_s=(4)/(6)\cdot\pi\cdot r^3=(4)/(6)\cdot3.14\cdot(1.3)^3=4.6cm^3 \\ V=V_t+V_c+V_s \\ V=2.12+9.55+4.6 \\ V=16.27cm^3 \end{gathered}`

The volume of the upper container is 16.27 cm^3, and being symmetrical, it is the same for the bottom container.

At the moment that all the sand finishes going to the bottom container, the height will be the sum of the heights in each case.

Then:

`\begin{gathered} h=1.2+1.8+1.3 \\ h=4.3\text{ cm} \end{gathered}`

Therefore, the height is 4.3 centimeters

To reach that height, all the sand had to be passed from one side to the other, therefore, we can calculate the time as follows:

`\begin{gathered} t=(16.27cm^3)/(0.05(cm^3)/(s)) \\ t=325.4\text{ sec} \end{gathered}`

It would take a time of 325.4 seconds